Debugging an FFT...

If an FFT time-domain input comprises say, 1.5 cycles of sine starting 
positive going followed by another 1.5 cycles of sine starting positive 
going - in other words has 3 cycles with an inversion half way along, 
then the real and imaginary FFT outputs at the 3-cycle frequency should 
be zero.

That's right, isn't it?

Cheers
-- 
Syd

On Thu, 19 Jul 2012 09:46:53 +0100, Syd Rumpo <u...@neonica.co.uk>
wrote:

>Debugging an FFT...
>
>If an FFT time-domain input comprises say, 1.5 cycles of sine starting 
>positive going followed by another 1.5 cycles of sine starting positive 
>going - in other words has 3 cycles with an inversion half way along, 
>then the real and imaginary FFT outputs at the 3-cycle frequency should 
>be zero.
>
>That's right, isn't it?
>
>Cheers
>-- 
>Syd
>

No.

http://www.dsprelated.com/showarticle/174.php


Eric Jacobsen
Anchor Hill Communications
www.anchorhill.com

On Thu, 19 Jul 2012 09:46:53 +0100, Syd Rumpo <u...@neonica.co.uk>
wrote:

>Debugging an FFT...
>
>If an FFT time-domain input comprises say, 1.5 cycles of sine starting 
>positive going followed by another 1.5 cycles of sine starting positive 
>going - in other words has 3 cycles with an inversion half way along, 
>then the real and imaginary FFT outputs at the 3-cycle frequency should 
>be zero.
>
>That's right, isn't it?
>
>Cheers
>-- 
>Syd

I shouldn't have been so terse with my previous response, and I
actually misinterpreted what you'd said, anyway.

It does look like bin 3 should be zero. 

The computation for bin 3 is essentially a correlator, and since the
correlations for the first half and the second half will have opposite
sign they cancel.   The discontinuity adds a lot of other frequency
content, though, that gets splattered around among the other bins.


Eric Jacobsen
Anchor Hill Communications
www.anchorhill.com

On 19/07/2012 17:06, Eric Jacobsen wrote:
> On Thu, 19 Jul 2012 09:46:53 +0100, Syd Rumpo <u...@neonica.co.uk>
> wrote:
>
>> Debugging an FFT...
>>
>> If an FFT time-domain input comprises say, 1.5 cycles of sine starting
>> positive going followed by another 1.5 cycles of sine starting positive
>> going - in other words has 3 cycles with an inversion half way along,
>> then the real and imaginary FFT outputs at the 3-cycle frequency should
>> be zero.
>>
>> That's right, isn't it?
>>
>> Cheers
>> --
>> Syd
>
> I shouldn't have been so terse with my previous response, and I
> actually misinterpreted what you'd said, anyway.
>
> It does look like bin 3 should be zero.
>
> The computation for bin 3 is essentially a correlator, and since the
> correlations for the first half and the second half will have opposite
> sign they cancel.   The discontinuity adds a lot of other frequency
> content, though, that gets splattered around among the other bins.
>
>
> Eric Jacobsen
> Anchor Hill Communications
> www.anchorhill.com

Thanks, Eric.

-- 
Syd